What is the area of the region enclosed by the curves y=lnx and y=ln^2x ?

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First, we'll determine the limits of integration. These limits are represented by the intercepting points of the given curves.

We'll equate:

(ln x)^2 = ln x

We'll subtract ln x:

(ln x)^2 - ln x = 0

We'll factorize by ln x:

ln x*(ln x - 1) = 0

We'll cancel each factor:

ln x = 0 => x = e^0 = 1

ln x - 1 = 0 => ln x = 1 => x = e

The intercepting points are (1 , 0) and (e , 1).

The lower limit of integration is x = 0 and the upper limit of integration is x = e.

To determine what curve is above of the other, we'll determine the monotony of the first derivative of the function f(x) = (ln x)^2 - ln x.

f'(x) = 2lnx/x - 1/x

f'(x) = (2ln x - 1)/x

If x = 1 => f'(1) = -1

If x = e => f'(e) = 1/e

The curve that is above is ln x and the area of the region is the definite integral of the function: ln x - (ln x)^2.

Int [ln x - (ln x)^2]dx = Int ln x dx - Int (ln x)^2 dx

We'll calculate Int ln x dx by parts:

Int udv = uv - INt vdu

Let u = ln x => du = dx/x

Let dv = dx => v = x

Int ln x dx = x*ln x - Int dx

Int ln x dx = x*(ln x - 1)

We'll apply Leibniz Newton to determine the definite integral:

Int ln x dx = F(e) - F(1)

F(e) = e*(ln e - 1) = e*(1-1) = 0

F(1) = -1

F(e) - F(1) = 1

We'll calculate Int (ln x)^2 dx by parts:

Let u = (ln x)^2 => du = 2ln xdx/x

Let dv = dx => v = x

Int (ln x)^2dx = x*(ln x)^2 - Int 2ln xdx

Int (ln x)^2dx = x*(ln x)^2 - 2*Int ln xdx

But Int ln x dx = 1

Int (ln x)^2dx = x*(ln x)^2 - 2

We'll apply Leibniz Newton to determine the definite integral:

F(e) - F(1) = e - 2

Int [ln x - (ln x)^2]dx = 1 - e + 2

Int [ln x - (ln x)^2]dx = 3 - e

**The area of the region bounded by the given curves is of (3 - e) square units.**

First we need to determine the points of intersection between lnx and ln^2 x

==> ln x = ln^2 x

==> ln^2 x - ln x = 0

==> lnx ( lnx -1) =0

==> lnx = 0 ==> x = 1

==> lnx-1 = 0 ==> lnx = 1 ==> x = e

Then we need to find the integral between x = 1 and x= e

==> Int ln^2 x - lnx = Int lnx ( lnx-1) dx

Int lnx dx

v= lnx ==> dv = 1/x dx

du = dx ==> u = x

==> Int vdu = v*u - Int u dv

==> Int lnx dx = xlnx - Int x*1/x dx

==> int lnx dx = xlnx - x + C........(1)

==> Int ln^2 x

u= ln^2x ==> du = 2lnx 1/x dx

dv = dx ==> v = x

==> Int u dv = u*v - Int v du

==> Int ln^2 x dx = xln^2 x - Int 2lnx dx

==> Int ln^2 x = xln^2 x - 2* (1)

==> Int ln^2 x = xln^2 x - 2( xlnx -x) + C

==> Int ln^2 x = xln^2 x - 2xlnx + 2x + C .............(2)

Now we will find the definite integral for both (1) and (2).

==> In lnx dx ( x= 1 tp x= e) = xlnx -x ( x=1 tp x= e)

==> elne - e - ( 1ln 1 - 1) = e - e - 0 + 1 = 1

Then the area under y= lnx is 1 square unit.

==> Int ln^2 x dx ( x= 1 tp x = e)

==> xln^2 x -2xlnx + 2x

==> ( eln^2 e - 2elne + 2e - ( 1ln^2 1 - 2ln1 + 2)

==> e - 2e + 2e - 2 = e-2

Then the area under the curve ln^2 x is e-2 square units.

Then the area between lnx and ln^2 x is e-2-1 = l e-3l = 3- e

**Then the area between the curves is (3-e) square units.**

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